Question

Let $\mathrm{P}(\mathrm{r})\left[\mathrm{Q} /\left(\pi \mathrm{R}^{4}\right)\right] \mathrm{r}$ be the charge density distribution for a solid sphere of radius \(\mathrm{R}\) and total charge \(\mathrm{Q}\). For a point ' \(\mathrm{P}\) ' inside the sphere at distance \(\mathrm{r}_{1}\) from the centre of the sphere the magnitude of electric field is (A) $\left[\mathrm{Q} /\left(4 \pi \epsilon_{0} \mathrm{r}_{1}^{2}\right)\right]$ (B) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(4 \pi \in{ }_{0} \mathrm{R}^{4}\right)\right]$ (C) $\left[\left(\mathrm{Qr}_{1}^{2}\right) /\left(3 \pi \epsilon_{0} \mathrm{R}^{4}\right)\right]$

Short answer

The short answer is: (B) \[\frac{Qr_{1}^2}{4\pi \epsilon_{0} R^4}\]

Step-by-step solution

Gaussian Surface

To apply Gauss's law for a solid sphere with charge density distribution, we need to choose an appropriate Gaussian surface. In this case, we will choose a sphere with radius r₁ centered at the center of the charged sphere because of the symmetry.

Setup Gauss's Law

Gauss's law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε₀). Mathematically, it's given by: \[\oint \vec{E} \cdot d\vec{A} = \frac{Q_\text{enclosed}}{\epsilon_{0}}\] Where \(\vec{E}\) is the electric field vector, \(d\vec{A}\) is the area vector of the Gaussian surface, and \(Q_\text{enclosed}\) is the total charge enclosed by the Gaussian surface.

Calculate Enclosed Charge

To find the enclosed charge, we need to integrate the charge density over the volume enclosed by our Gaussian surface (a sphere with radius r₁): \[Q_\text{enclosed} = \int_{V} P(r) dV\] Where \(P(r) = [\frac{Q}{\pi R^4}]r\) and \(dV = 4\pi r^2 dr\). Integrate over the volume of the Gaussian sphere (from 0 to r₁): \[Q_\text{enclosed} = \int_{0}^{r_{1}} \left[\frac{Q}{\pi R^4}\right]r(4\pi r^2 dr)\]

Solve Integral

Performing the integration, we get: \[Q_\text{enclosed} = \frac{4Q}{R^4}\int_{0}^{r_{1}} r^3 dr = \frac{4Q}{R^4}\left[\frac{r_{1}^4}{4}\right] = \frac{Qr_{1}^4}{R^4}\]

Apply Gauss's Law

Now, we can apply Gauss's law for our Gaussian sphere: \[\oint \vec{E} \cdot d\vec{A} = \frac{Q_\text{enclosed}}{\epsilon_{0}}\] Due to the spherical symmetry, the electric field has the same magnitude at every point on the Gaussian surface and is parallel to the area vector. Thus, their dot product simplifies to their magnitudes multiplied together: \[\oint E dA = E\oint dA = E \cdot 4\pi r_{1}^2 = \frac{Q_\text{enclosed}}{\epsilon_{0}}\] Let's substitute the value of \(Q_\text{enclosed}\) from Step 4: \[E \cdot 4\pi r_{1}^2 = \frac{Qr_{1}^4}{R^4\epsilon_{0}}\]

Solve for Electric Field Magnitude

Finally, we solve for the electric field magnitude E: \[E = \frac{Qr_{1}^2}{4\pi R^4\epsilon_{0}}\] Now let's compare this result with the given options. We can see that our result matches option (B), so the correct answer is: (B) \[\frac{Qr_{1}^2}{4\pi \epsilon_{0} R^4}\]